Let $S$ be a surface in 3D described by the equation $3x + \sin(y) + z^2 = 0$. What is the equation of the plane tangent to $S$ at $(-3, \pi, 3)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $2(z - 6) = 0$ (Choice B) B $3(x - 3) - (y + \pi) + 6(z + 3) = 0$ (Choice C) C $-3(x + 3) - \pi(y - 1) + 3(z - 6) = 0$ (Choice D) D $3(x + 3) - (y - \pi) + 6(z - 3) = 0$
The equation for a tangent plane of an implicitly defined surface $F(x, y, z) = 0$ at the point $(a, b, c)$ is: $F_x(x - a) + F_y(y - b) + F_z(z - c) = 0$ [What's the intuition behind the formula?] Let's find $F_x$, $F_y$, and $F_z$. $\begin{aligned} F_x &= 3 \\ \\ F_y &= \cos(y) = -1 \\ \\ F_z &= 2z = 6 \end{aligned}$ Putting it all together, here's the equation for the tangent plane of $S$ at $(-3, \pi, 3)$ : $3(x + 3) - (y - \pi) + 6(z - 3) = 0$